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The problem in the image asks to **"Compute the value of \( \sum_{n=1}^{100} \left( 2^n - 3^n \right) \), given \( \sum_{n=1}^{k} x^n = \frac{x(x^k - 1)}{x - 1} \) for any integer \( k \geq 1 \) and \( x \neq 1 \)."**

### Solution:
We are required to compute the sum:
\[
\sum_{n=1}^{100} \left( 2^n - 3^n \right)
\]
This can be split into two separate sums:
\[
\sum_{n=1}^{100} 2^n - \sum_{n=1}^{100} 3^n
\]
Let's compute each sum separately using the formula provided:
\[
\sum_{n=1}^{k} x^n = \frac{x(x^k - 1)}{x - 1}
\]
For \( x = 2 \) and \( k = 100 \):
\[
\sum_{n=1}^{100} 2^n = \frac{2(2^{100} - 1)}{2 - 1} = 2(2^{100} - 1) = 2^{101} - 2
\]
For \( x = 3 \) and \( k = 100 \):
\[
\sum_{n=1}^{100} 3^n = \frac{3(3^{100} - 1)}{3 - 1} = \frac{3(3^{100} - 1)}{2} = \frac{3 \times 3^{100} - 3}{2} = \frac{3^{101} - 3}{2}
\]
Thus, the value of the original sum is:
\[
\sum_{n=1}^{100} \left( 2^n - 3^n \right) = \left( 2^{101} - 2 \right) - \frac{3^{101} - 3}{2}
\]
Let's simplify this expression:
\[
= 2^{101} - 2 - \frac{3^{101}}{2} + \frac{3}{2}
\]
Finally, the simplified expression is:
\[
\boxed{2^{101} - \frac{3^{101}}{2} - 2 + \frac{3}{2}}
\]

Do you want any further details on this problem or have any questions?

Here are five related questions for practice:

1. How would the solution change if the upper limit was 50 instead of 100?
2. What would be the sum if \( x = 4 \) instead of 2 in the given summation formula?
3. Can you derive the general form for \( \sum_{n=1}^{k} x^n \) using the geometric series formula?
4. What happens to the sum if \( x = 1 \)?
5. How does the sum behave as \( k \) approaches infinity?

**Tip:** When dealing with sums involving powers, always check if a geometric series formula or a known summation identity can simplify the calculation.

Compute the value of \( \sum_{n=1}^{100} \left( 2^n - 3^n \right) \), given \( \sum_{n=1}^{k} x^n = \frac{x(x^k - 1)}{x - 1} \) for any integer \( k \geq 1 \) and \( x \neq 1 \).

Explore how to compute the summation \( \sum_{n=1}^{100} \left( 2^n - 3^n \right) \) using the geometric series sum formula. This problem involves exponential functions and is suitable for advanced high school students (Grades 11-12).

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